1.

Find the vector equation of the plane passing through the point (2,1,-1) and normal to the plane is \(2\hat{i}+\hat{j}-3\hat{k}\)?(a) \((\vec{r}-(2\hat{i}-7\hat{k})).(2\hat{i}+\hat{j}-3\hat{k})\)=0(b) \((\vec{r}-(2\hat{i}+3\hat{j}-\hat{k})).(2\hat{i}-3\hat{k})\)=0(c) \((\vec{r}-(\hat{i}+\hat{j}-3\hat{k})).(2\hat{i}+6\hat{j}-3\hat{k})\)=0(d) \((\vec{r}-(2\hat{i}+\hat{j}-\hat{k})).(2\hat{i}+\hat{j}-3\hat{k})\)=0The question was asked in an online interview.Query is from Three Dimensional Geometry topic in portion Three Dimensional Geometry of Mathematics – Class 12

Answer»

Correct OPTION is (d) \((\vec{R}-(2\HAT{i}+\hat{j}-\hat{k})).(2\hat{i}+\hat{j}-3\hat{k})\)=0

To elaborate: The position VECTOR of the point (2,1,-1) is \(\vec{a}=2\hat{i}+\hat{j}-\hat{k}\) and the normal vector \(\vec{N}\)perpendicular to the plane is \(\vec{N}=2\hat{i}+\hat{j}-3\hat{k}\)

The vector equation of the plane is given by \((\vec{r}-\vec{a}).\vec{N}\)=0

Therefore, \((\vec{r}-(2\hat{i}+\hat{j}-\hat{k})).(2\hat{i}+\hat{j}-3\hat{k})\)=0


Discussion

No Comment Found

Related InterviewSolutions